Given $X$ is a random variable with uniform distribution over $(-2, 1)$ and $Y=X^4$, where $Y$ is also a random variable. Find distribution function $F_Y$ and pdf $f_Y$.
I am using the monotonic (inverse) transformation method with $x= \pm\sqrt[4]y$.
So$$ F_Y(y)= F_X(\sqrt[4]y) – F_X(-\sqrt[4]y)\implies F_Y(y)= \dfrac23\sqrt{y}.$$
Since the value of $x$ is in $(-2,1)$ and that of $y$ in $(0,16)$, my final answer was$$
F_Y(y) = \begin{cases}
\dfrac23\sqrt{y}; & 0< y <16\\
\dfrac16\sqrt[4]{y^{-3}}; &0 < y < 16
\end{cases}
$$
The answer was only given for $f_Y(y)$ and it was$$
f_Y(y) = \begin{cases}
\dfrac16\sqrt[4]{y^{-3}}; & 0 < y < 1\\
\dfrac1{12}\sqrt[4]{y^{-3}}; &1 < y < 16
\end{cases}
$$
Any idea where I went wrong?
Best Answer
You can't use the inverse transform method because the mapping $X^4$ is not monotone. You can instead apply the monotone transform theorem to the sets where it is monotone, namely $(-2,0)$ and $(0,1)$.
Specifically, $X = g^{-1}(Y)$ is a well defined inverse on $(-2,1)$ and $(0,1)$ respectively. The Jacobian of this transformation is $\left|\frac{d}{dy}g^{-1}(y)\right| = \frac{1}{4}y^{-3/4}$. So, the pdf of $Y$ is given by,
$$ f_Y(y) = \frac{1}{12}y^{-3/4}\chi_{(0,16)}(y) + \frac{1}{12}y^{-3/4}\chi_{(0,1)}(y) = \frac{1}{12}y^{-3/4}\chi_{(1,16)}(y)+\frac{1}{6}y^{-3/4}\chi_{(0,1)}(y)$$